Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1993 AHSME Problems/Problem 7"
(→‎Solution)
Line 10: Line 10:
  
 
== Solution ==
 
== Solution ==
−
Because the only digits we are using are 1s and zeroes (they say that in question) we essentially perform the operation in binary and convert back to base 10 to get the answer.
+
Note <math>R_n = \sum_{k=0}^{n-1} 10^k = \frac{10^n - 1}{10-1}</math>.
−
Notice then that <math>R(n)_2=(2^n-1)_{10}</math>. It follows that <math>R(24)=2^{24}-1</math> and <math>R(4)=2^4-1</math>
+
 
−
Notice to compute <math>\frac{2^{24}-1}{2^4-1}</math> we take advantage of the fact that
+
Therefore <math>\frac{R_{24}}{R_4} = \frac{ 10^{24}-1 }{10^4-1}</math>. 
−
<math>x^6-1=(x-1)(x^5+x^4+x^3+x^2+x+1)</math>
+
 
−
Our quotient then is just
+
But we can recognize this form as the sum of a geometric series <math>1+10^4 + (10^4)^2 + \dots + (10^4)^5 = 1+ 10^4 + 10^8 + \dots + 10^{20}</math>
−
<math>2^{20}+2^{16}+2^{12}+2^{8}+2^4+1</math>
+
 
−
Notice then this just a <math>21</math> digit in binary with <math>5</math> <math>1</math>s which occupy the <math>2^a</math> slots for the <math>6</math> <math>a</math> we have.
+
Now the 1's place has a 1, but the 10's, 100's and 1,000's place have 0's.  The 10,000's place has a 1, but the <math>10^5</math>, <math>10^6</math> and <math>10^7</math> places have 0's. Between successive 1's in the decimal expansion, there are three 0's, which gives <math>5\times 3=15</math> zeros altogether.
−
Our answer then is just <math>21-6=\boxed{15}</math>
+
 
−
<math>\fbox{E}</math>
+
The answer is <math>\fbox{E}</math>
  
 
== See also ==
 
== See also ==

Revision as of 20:27, 27 May 2021

Problem

The symbol $R_k$ stands for an integer whose base-ten representation is a sequence of $k$ ones. For example, $R_3=111,R_5=11111$, etc. When $R_{24}$ is divided by $R_4$, the quotient $Q=R_{24}/R_4$ is an integer whose base-ten representation is a sequence containing only ones and zeroes. The number of zeros in $Q$ is:

$\text{(A) } 10\quad \text{(B) } 11\quad \text{(C) } 12\quad \text{(D) } 13\quad \text{(E) } 15$

Solution

Note $R_n = \sum_{k=0}^{n-1} 10^k = \frac{10^n - 1}{10-1}$.

Therefore $\frac{R_{24}}{R_4} = \frac{ 10^{24}-1 }{10^4-1}$.

But we can recognize this form as the sum of a geometric series $1+10^4 + (10^4)^2 + \dots + (10^4)^5 = 1+ 10^4 + 10^8 + \dots + 10^{20}$

Now the 1's place has a 1, but the 10's, 100's and 1,000's place have 0's. The 10,000's place has a 1, but the $10^5$, $10^6$ and $10^7$ places have 0's. Between successive 1's in the decimal expansion, there are three 0's, which gives $5\times 3=15$ zeros altogether.

The answer is $\fbox{E}$

See also

1993 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1993_AHSME_Problems/Problem_7&oldid=154333"